Experiments Related to Units and Measurements, Vernier calipers-its use to measure the internal and external diameter and depth of a vessel. Screw gauge-its use to determine thickness/ diameter of thin sheet/wire. Metre Scale - the mass of a given object by the principle of moments.: NEET MCQs with Answers

8 free questions on Experiments Related to Units and Measurements, Vernier calipers-its use to measure the internal and external diameter and depth of a vessel. Screw gauge-its use to determine thickness/ diameter of thin sheet/wire. Metre Scale - the mass of a given object by the principle of moments. for NEET (UG). Try each one, then tap Show answer for the answer and a short explanation.

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Question 1

A student measures the diameter of a wire with a screw gauge whose least count is 0.01 mm. Which of the following is the correct way to record the reading?
  1. 2.4 mm
  2. 2.40 mm
  3. 2.400 mm
  4. 2.4000 mm
Show answer
Answer: (B) 2.40 mm
Explanation: A reading should be recorded up to the least count of the instrument. With a least count of 0.01 mm, the reading must have two decimal places in millimetres, so 2.40 mm is correct.

Question 2

In a vernier callipers, N divisions of the main scale coincide with (N + 1) divisions of the vernier scale. If one main scale division is 1 mm, the least count of the instrument is:
  1. N/(N + 1) mm
  2. 1/(N + 1) mm
  3. 1/N mm
  4. 2/(N + 1) mm
Show answer
Answer: (B) 1/(N + 1) mm
Explanation: (N + 1) vernier divisions equal N mm, so one vernier division is N/(N + 1) mm. Least count = 1 MSD − 1 VSD = 1 − N/(N + 1) = 1/(N + 1) mm.

Question 3

A micrometer screw gauge has a pitch of 2 mm and 200 divisions on its circular scale. With the jaws closed it shows a positive zero error, its circular scale reading being 6 divisions. With a thin sheet between the jaws, the main scale shows 2 divisions (each 2 mm) and the circular scale reads 73 divisions. The thickness of the sheet is
  1. 4.67 mm
  2. 2.67 mm
  3. 4.73 mm
  4. 4.79 mm
Show answer
Answer: (A) 4.67 mm
Explanation: Least count = 2/200 = 0.01 mm, so the zero error is +0.06 mm. The observed reading is 2 × 2 + 73 × 0.01 = 4.73 mm. The corrected thickness is 4.73 − 0.06 = 4.67 mm.

Question 4

A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. With the jaws closed, the 3rd division of the circular scale lies on the reference line while the zero of the circular scale is below it (positive zero error). With a small ball between the jaws, the main scale reading is 6 divisions (each 0.5 mm) and the 32nd circular scale division lies on the reference line. The corrected diameter of the ball is
  1. 3.29 mm
  2. 3.32 mm
  3. 3.35 mm
  4. 6.29 mm
Show answer
Answer: (A) 3.29 mm
Explanation: Least count = 0.5/50 = 0.01 mm, so the zero error is +3 × 0.01 = +0.03 mm. The observed reading is 6 × 0.5 + 32 × 0.01 = 3.32 mm. Subtracting the positive zero error gives 3.32 − 0.03 = 3.29 mm.

Question 5

A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale, and no zero error. For a wire it shows 1 mm on the pitch scale and the 36th circular division on the reference line. The wire's length, measured with a metre scale, is 8.0 cm. What are its diameter and its volume, the volume being stated to the correct number of significant figures?
  1. 1.36 mm; 0.12 cm³
  2. 1.72 mm; 0.19 cm³
  3. 1.36 mm; 0.116 cm³
  4. 1.36 mm; 0.1162 cm³
Show answer
Answer: (A) 1.36 mm; 0.12 cm³
Explanation: The least count is 0.5 mm/50 = 0.01 mm, so the diameter is 1 + 36 × 0.01 = 1.36 mm. The volume is (π/4)(0.136 cm)²(8.0 cm) = 0.1162 cm³ by calculation. The length has only two significant figures, so the result must be stated to two significant figures, as 0.12 cm³.

Question 6

The jaws of a vernier callipers are closed with nothing between them and the zero of the vernier scale does not coincide with the zero of the main scale. Which description is correct?
  1. Vernier zero to the right of main zero: negative zero error; readings are too small, so the magnitude of the error is added to them
  2. Vernier zero to the left of main zero: negative zero error; readings are too large, so the magnitude of the error is subtracted from them
  3. Vernier zero to the left of main zero: positive zero error; readings are too small, so the error is subtracted from them
  4. Vernier zero to the right of main zero: positive zero error; readings are too large, so the error is subtracted from them
Show answer
Answer: (D) Vernier zero to the right of main zero: positive zero error; readings are too large, so the error is subtracted from them
Explanation: If the vernier zero lies to the right, the instrument already shows a positive length with the jaws closed, so the zero error is positive and every reading is too large. If it lies to the left, the zero error is negative and every reading is too small. In both cases the zero error is subtracted algebraically from the observed reading, which adds the magnitude when the error is negative.

Question 7

A screw gauge has a pitch of 0.5 mm and 50 divisions on its circular scale. It gives the sides of a small rectangular plate as 8.00 mm and 2.00 mm. Taking the error in each reading as one least count, what is the maximum percentage error in the calculated area of the plate?
  1. 0.375%
  2. 0.625%
  3. 1.25%
  4. 0.5%
Show answer
Answer: (B) 0.625%
Explanation: The least count is 0.5 mm/50 = 0.01 mm. The percentage errors in the sides are (0.01/8.00) × 100 = 0.125% and (0.01/2.00) × 100 = 0.5%. For a product, the maximum percentage errors add, giving 0.625% for the area.
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Question 8

A special vernier calliper has a main scale marked in millimetres, and its 10 vernier divisions span 11 main scale divisions. It has no zero error. In a measurement, the vernier zero lies between the 3.2 cm and 3.3 cm marks, and the 4th vernier division coincides with a main scale mark. The reading is
  1. 3.16 cm
  2. 3.24 cm
  3. 3.28 cm
  4. 3.26 cm
Show answer
Answer: (D) 3.26 cm
Explanation: Each vernier division here is 1.1 mm long. If the vernier zero is at 32 mm + f, the 4th division is at 32 + f + 4.4 mm, and this must fall on a whole millimetre mark, which requires f = 0.6 mm. The reading is therefore 32.6 mm = 3.26 cm. As a check, 32.6 + 4.4 = 37.0 mm lies exactly on a main scale mark.

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